Problem 14.3b3

A First Course in String Theory
 
 
14.3 Massive level in the open superstring.
 
~~~
 
What is a zero mode?
 
p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states.”
 
 
How come there are creation operators that do not contribute to the mass-squared of the states?
 
Before each creation operator, there is a multiple which is the same as the absolute value of the index of the creation operators.
 
If an index can be zero, the corresponding term can be zero.
 
— Me@2015.09.26 08:44 PM
 
 
Consider the NS-sector:
 
Equation (14.37):
 
M^2 = \frac{1}{\alpha'} \left( \frac{-1}{2} + N^\perp \right)
 
N^\perp = \sum_{p=1}^\infty \alpha_{-p}^I \alpha_p^I + \sum_{r=\frac{1}{2}, \frac{3}{2} ...} r b_{-r}^I b_r^I
 
For the NS sector, the r values in b^I_{r} are half-integers, thus cannot be zero. So every creation operator b_{-r}^I contributes to the mass-squared.
 
(p.312 “… the negatively moded coefficients b_{-1/2}^I, b_{-3/2}^I, b_{-5/2}^I, …, are creation operators, …”)
 
 
Consider the R-sector:
 
Equation (14.53):

M^2 = \frac{1}{\alpha'} \sum_{n \ge 1} \left( \alpha_{-n}^I \alpha_{n}^I + n d_{-n}^I d_n^I \right)
 
For the R sector, the n values in d^I_n are integers, thus can be zero. So some of the creation operators d^I_{-n} are zero modes.
 
p.315 “… the eight fermionic zero modes d_0^I…”
 
— Me@2015-10-11 11:01:44 AM
 
 
 
2015.10.11 Sunday (c) All rights reserved by ACHK